John Travis
Mississippi College
Playing around with solving first order linear Differential Equations.
We will presume the DEs are in the form $\frac{dy}{dt} = P(t) y + Q(t)$
reset()
|
|
# Solving a first order linear DE in general using Sage
var('t')
y = function('y',t)
P = function('P',t)
Q = function('Q',t)
DE = diff(y,t)==P*y+Q
soln=desolve(DE,y,ivar=t)
show(soln)
|
\newcommand{\Bold}[1]{\mathbf{#1}}{\left(c + \int e^{\left(-\int P\left(t\right)\,{d t}\right)} Q\left(t\right)\,{d t}\right)} e^{\left(\int P\left(t\right)\,{d t}\right)}
\newcommand{\Bold}[1]{\mathbf{#1}}{\left(c + \int e^{\left(-\int P\left(t\right)\,{d t}\right)} Q\left(t\right)\,{d t}\right)} e^{\left(\int P\left(t\right)\,{d t}\right)}
|
# Now, let's try specifying values for P(t) and Q(t) and see what happens
var('t,c')
t0 = 0 # starting point for time
t1 = 3 # ending point for time
y0 = 1 # starting y-value
y = function('y',t)
P = function('P',t)
Q = function('Q',t)
P = -1/2
Q = 2*cos(t)
DE = diff(y,t)==P*y+Q
soln=desolve(DE,y,ivar=t,ics=[t0,y0])
show(soln.expand())
|
|
G=points(desolve_rk4(DE, y, ics=[t0,y0], ivar=t, end_points=3, step=0.25))
G+=plot(soln,t,[t0,t1])
show(G)
|
show(integrate(2*cos(t)*e^(-1/2*t),t).expand())
|
|
var('w')
Plot1=plot_slope_field(1/2*w+2*cos(t),(t,0,10),(w,-6,6))+plot(soln,t,0,3,color='red',thickness=5)+G
Plot1.show()
|
|
|